HDOJ 1260 DP

Tickets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4393    Accepted Submission(s): 2236

Problem Description

Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible.
A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time.
Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help.

 

Input

There are N(1<=N<=10) different scenarios, each scenario consists of 3 lines:
1) An integer K(1<=K<=2000) representing the total number of people;
2) K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;
3) (K-1) integer numbers(0s<=Di<=50s) representing the time needed for two adjacent people to buy two tickets together.

 

Output

For every scenario, please tell Joe at what time could he go back home as early as possible. Every day Joe started his work at 08:00:00 am. The format of time is HH:MM:SS am|pm.

 

Sample Input

2 2 20 25 40 1 8

 

Sample Output

08:00:40 am 08:00:08 am

 

Source

浙江工业大学第四届大学生程序设计竞赛

题意：

知道一个人买票花的时间和和前面那个人一起买票花的时间，问最少花多少时间可以把票卖完..

输入：

　　　　给出T，表示有T组样例

　　　　给出n，表示有n个人买票..

　　　　给出n个数表示这个人单独买票会花的时间..

　　　　给出n-1个数，表示这个人和前面那个人一起买票会花的时间..

思路：dp[i]表示前i个人所用的最短时间，则转移方程为：dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i-1]);

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<vector>
#define ll long long
#define  db double
using namespace std;
const int N=1e6+5;
const int mod=1e9+7;
int a[2005],b[2005];
ll dp[N];
int h,m,s;
int main(){
    int n;
    int k;
    scanf("%d",&n);
    while(n--){
        scanf("%d",&k);
        for(int j=1;j<=k;j++) scanf("%d",a+j);
        for(int j=1;j<=k-1;j++) scanf("%d",b+j);
        memset(dp,0, sizeof(dp));
        dp[1]=a[1];
        for(int i=2;i<=k;i++){
            dp[i]=min<ll>(dp[i-1]+a[i],dp[i-2]+b[i-1]);
        }
        h=dp[k]/3600+8,m=dp[k]%3600/60,s=dp[k]%60;
        if(h>12){
            printf("%02d:%02d:%02d pm
",h-12,m,s);
        }
        else  printf("%02d:%02d:%02d am
",h,m,s);
    }
}